Advanced Mathematical Methods for Scientists and Engineers Episode 3 Part 5 pdf

... make this an exact equation.ddxex 3 /3 y= c1ex 3 /3 ex 3 /3 y = c1ex 3 /3 dx + c2y = c1e−x 3 /3 ex 3 /3 dx + c2e−x 3 /3 9 45 Exercise 17.21Find the general solution ... real and positive.944Method 1. Note that this is an exact equation.ddx(y− x2y) = 0y− x2y = c1ddxe−x 3 /3 y= c1e−x 3 /3 y = c1ex 3 /3 e−x 3 /3 dx + c2ex 3 /3 Method ... Solution 954 Noting thate(2+3i)x=e2x[cos(3x) + ı sin(3x)]e(2−3i)x=e2x[cos(3x) − ı sin(3x)],we can write the two linearly independent solutionsy1=e2xcos(3x), y2=e2xsin(3x).Solution...

Advanced Mathematical Methods for Scientists and Engineers Episode 1 Part 5 pdf

... +√872 /3 − 6−1 /3 9 +√871 /3 2≈ (0 .58 9 755 , 0, 0 .34 781).The closest point is shown graphically in Figure 5. 10.-1-0 .5 00 .5 1-1-0 .5 00 .5 100 .5 11 .5 2-1-0 .5 00 .5 100 .5 11 .5 2Figure ... =6−2 /3 9 +√872 /3 − 6−1 /3 9 +√871 /3 ≈ 0 .58 9 755 .Thus the closest point to (1, 0, 0) on the paraboloid is6−2 /3 9 +√872 /3 − 6−1 /3 9 +√871 /3 , 0,6−2 /3 9 ... dx=2 3 x 3/ 210+2 3 x 3/ 220=2 3 +2 3 2 3/ 2=2 3 (1 + 2√2)Solution 4 .3 ddxx2xf(ξ) dξ = f(x2)ddx(x2) − f(x)ddx(x)= 2xf(x2) − f(x) 151 12 3 4 5 -101-10112 3 4 5 -101Figure...

Advanced Mathematical Methods for Scientists and Engineers Episode 4 Part 5 pdf

... sin(nπx) dx=4(nπ)2sin(nπ/2)=4(nπ)2(−1)(n−1)/2 for odd n0 for even n. 1 35 6 -3 -2-1 12 3 -1 .5 -1-0 .5 0 .5 11 .5 Figure 28 .5: Fourier Sine Series.The functions {. . . ,e−ıx, 1,eıx,eı2x, ... -1-0 .5 0 .5 1-1-0 .5 0 .5 1-1-0 .5 0 .5 1-0.4-0.20.20.4Figure 28.7: Three Term Approximation for a Function with Jump Discontinuities and a Continuous Function.A ... 4∞n=1oddncos(nx). For x = nπ, this implies0 = 4∞n=1oddncos(nx), 136 1-1-0 .5 0 .5 1-0.2-0.10.10.210. 25 0.100.11110 .5 Figure 28.8: Three Term Approximation for a Function with...

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Advanced Mathematical Methods for Scientists and Engineers Episode 6 Part 5 pdf

... areλ1≈ 0.678298λ2≈ 7.27 931 λ 3 ≈ 24. 930 2λ4≈ 54 . 25 93 λ 5 ≈ 95 .30 57 The eigenfunctions are,φn(x) = v(1; λn)u(x; λn) − u(1; λn)v(x; λn).Solution 48 .32 1. First note thatsin(kx) ... equation isφ(x) =a + bx for λ = 0a cos√λx+ b sin√λx for λ > 0a cosh√−λx+ b sinh√−λx for λ < 0We see that for λ = 0 and λ < 0 only the trivial solution ... (48.8)21 45 3. We consider the problemu − λT u = f. For λ = λ, (λ not an eigenvalue), we can obtain a unique solution for u.u(x) = f(x) +2π0Γ(x, s, λ)f(s) dsSince K(x, s) is self-adjoint and...

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Advanced Mathematical Methods for Scientists and Engineers Episode 1 Part 1 pdf

... . . . . 150 7 32 The Fourier Transform 1 53 9 32 .1 Derivation from a Fourier Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 53 9 32 .2 The Fourier Transform . . ... q(x).Hint, Solution12 33 The Gamma Function 16 05 33 .1 Euler’s Formul a . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16 05 33 .2 Hankel’s Formula . . . . ... . . . . . . . . 155 9 32 .5 Solving Diﬀerential Equations with the Fourier Transform . . . . . . . . . . . . . . . . . . . . . . . . 156 0 32 .6 The Fourier Cosine and Sine Transform . . . . . ....

Advanced Mathematical Methods for Scientists and Engineers Episode 2 Part 5 pps

... expansions of 1/(1 + z). For |z| < 1,11 + z= 1 +−11z +−12z2+−1 3 z 3 + ···= 1 + (−1)1z + (−1)2z2+ (−1) 3 z 3 + ···= 1 − z + z2− z 3 + ··· 55 8 For |z| > 1,11 ... is uniformly convergent for |z| ≤ r < 1. 54 512.2.2 Uniform Convergence and Continuous Functions.Consider a series f(z) =∞n=1an(z) that is uniformly convergent in s ome domain and whose ... n)n11.∞n=2(−1)nln1n12.∞n=2(n!)2(2n)! 13. ∞n=2 3 n+ 4n+ 5 5n− 4n− 3 56 2Im(z)Re(z)RR21Im(z)Re(z)RR21Cr1r2zCCC12zFigure 12 .5: Contours for a Laurent Expansion in an Annulus.Example...

Advanced Mathematical Methods for Scientists and Engineers Episode 2 Part 7 pdf

... 1))zn+ d, for 2 < |z|Solution 12.29The radius of convergence of the series for f(z) isR = limn→∞k 3 /3 k(k + 1) 3 /3 k+1= 3 limn→∞k 3 (k + 1) 3 = 3. 622The ... cosh zz 3 sin z sinh z=1 −z22+z424− ···1 +z22+z424+ ···z 3 z −z 3 6+z 5 120− ···z +z 3 6+z 5 120+ ···=1 −z46+ ···z 3 z2+ z6−1 36 +160+ ... 13. 1 .3 Evaluate the integralCcot z coth zz 3 dzwhere C is the unit circle about the origin in the positive direction.The integrand iscot z coth zz 3 =cos z cosh zz 3 sin z sinh z 632 Note...

Advanced Mathematical Methods for Scientists and Engineers Episode 3 Part 1 potx

... xnF1,yx.(Just formally substitute 1/x for λ.) For example,xy2,x2y + 2y 3 x + y, x cos(y/x)are homogeneous functions of orders 3, 2 and 1, respectively.Euler’s theorem for a homogeneous ... < 0, and deﬁne the solution, y, to bey(x) =y+(x), for x ≥ 0,y−(x), for x ≤ 0.The initial condition for y−demands that the solution be continuous.Solving the two problems for positive ... =−1 for x < 00 for x = 01 for x > 0.Since sign x is piecewise deﬁned, we solve the two problems,y++ y+= 0, y+(1) = 1, for x > 0y−− y−= 0, y−(0) = y+(0), for x...

Advanced Mathematical Methods for Scientists and Engineers Episode 3 Part 2 potx

... =1211e−t+121 5 e3t For large t, the solution looks likex ≈121 5 e3t.Both coordinates tend to inﬁnity.Figure 15. 1 shows some homogeneous solutions in the phase plane.Example 15. 2.2 (mathematica/ode/systems/systems.nb) ... the denominator of the fraction to see that z = 3 and z = − 3 are regular singular points.dwdz+1(z − 3) (z + 3) w = 0We make the transformation z = 1/ζ to examine the point at inﬁnity.dudζ−1ζ21(1/ζ)2+ ... A = −2 For positive A, the solution is bounded at the origin only for c = 0. For A = 0, there are no bounded solutions. For negative A, the solution is bounded there for any value of c and thus...

Advanced Mathematical Methods for Scientists and Engineers Episode 3 Part 3 pptx

... =2 0 10 2 00 1 3 2. Solvedxdt= Ax + g(t), x(0) = 0868 15. 6 HintsHint 15. 1Hint 15. 2Hint 15 .3 Hint 15. 4Hint 15. 5Hint 15. 6Hint 15. 7Hint 15. 8Hint 15. 9Hint 15. 10870where S is ... = 5 − λ 3 −28 5 − λ −4−4 3 3 −λ= −λ 3 + 3 2− 3 + 1 = −(λ − 1) 3 λ = 1 is an eigenvalue of multiplicity 3. The rank of the null space of A − I is 2. (The second and thirdrows ... =eAtx0= SeJtS−1x0=1 11 5 e−t00e3t14 5 −1−1 11 3 =12e−t+e3te−t +5 e3tx =1211e−t+121 5 e3tSolution 15. 5We consider an initial value...

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Advanced Mathematical Methods for Scientists and Engineers Episode 1 Part 1 pdf

Advanced Mathematical Methods for Scientists and Engineers Episode 3 Part 5 pdf